TITLE.PM5

406 ENGINEERING THERMODYNAMICS

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\M-therm\Th8-2.pm5

From compressibility charge, corresponding to pr = 0.093 and Tr = 9.1

Z (^) ~− 1
(This indicates that the gas having higher critical pressure and lower critical temperature
behaves like an ideal gas at normal pressure and temperature conditions.)
Also, pV = ZmRT
or m =
pV
ZRT =
1 21 10 904 78
1 8314
2
303
..××^5
×FHG IKJ×
= 86.9 kg. (Ans.)
Example 8.16. Determine the value of compressibility factor at critical point (Zcp) for the
Van der Waals’ gas.
Solution. Refer Fig. 8.13.
pcp
p
vcp v
C.P. = Critical point
Isotherms
C.P.
Fig. 8.13
From the isotherms plotted on p-v diagram in Fig. 8.13 it can be seen that the critical
isotherms has an inflection point, whose tangent is horizontal at the critical point.
∂
∂
p
v
c
cp
F
HG
I
KJ = 0 and
∂
∂
2
2
p
v cp = 0
The Van der Waal’s equation at the critical point is
pcp =
RT
vb
a
v
cp
cp cp
0
− −^2 ...(i)
As Tcp is constant
∂
∂
p
v
cp
cp
F
HG
I
KJ = − − +
RT
vb
a
v
cp
cp cp
0
23
2
()
= 0 ...(ii)
∂
∂
2
2
p
v
cp
cp
F
H
GG
I
K
JJ =^2 RT^0346
vb
a
v
cp
()()cp− cp
− = 0 ...(iii)